Torsten Sillke, BI, 199? Tiling the 3-cube with equal quasi-9-cubes: ------------------------------------------- The only possibility is the trivial 1x3x3 box. There are some cases to check. I didn't found a short proof so far. A packing with three congruent (reflections allowed) quasi-9-cubes is impossible too. As David Singmaster once said, that he had heard of another solution which needed reflection, I have redone my analysis by program testing all quasi-9-cubes fitting into a 2x3x3 box. This is sufficient if you consider the center cube. But no other quasi-9-cube packs the 3-cube. > Return-Path: > Date: Sat, 28 Mar 1998 20:57:27 -0500 > From: michael reid > Reply-To: nobnet@iijnet.or.jp > Subject: [NOBNET 791] angus lavery's challenge refuted > Sender: owner-nobnet@iijnet.or.jp > To: nobnet@iijnet.or.jp > Errors-To: owner-nobnet@iijnet.or.jp > Message-Id: <199803290157.KAA02789@ml0.iijnet.or.jp> > X-Sequence: NOBNET 791 > Precedence: bulk > Lines: 199 > > david singmaster writes > > > A somewhat different dissection problem for the 3x3x3 cube was posed > > by Angus Lavery some years ago, but I have never found a solution. The problem > > is to dissect the 3x3x3 into three congruent polycube pieces (so the unit > > cubes are not divided). Angus asserts it can be done, with one piece being a > > reflection of the others. > > > i think i can prove that the only solution is three 1x3x3 blocks, > even allowing disconnected shapes. this is a long and torturous > argument, and therefore somewhat susceptible to gaps and omissions. > > claim 1. the bounding box of such a shape cannot be 3x3x3. > proof. suppose the bounding box is 3x3x3. then either the shape > covers the central cell of the bounding box, or it doesn't. > in the first case, all three pieces cover the central cell > of the 3x3x3 cube. in the second case, none of the three > pieces cover the central cell. both are contradictions, > so the claim is proved. > > claim 2. the bounding box of such a shape is either 1x3x3 or 2x3x3. > proof. one of the three pieces must cover at least 3 of the 8 > corners of the cube. thus it must have width 3 in at least > two different directions. therefore the bounding box is > either 1x3x3 or 2x3x3, as claimed. > > if its bounding box is 1x3x3, then the shape is a 1x3x3 box. so we > only need to consider the case where the bounding box is 2x3x3. > label the cells of the bounding box as shown > > aba ded > bcb efe > aba ded > > note that there are two different positions of the bounding box inside > the 3x3x3 cube; they correspond to the "abc" layer being on an outer > layer of the cube, or the "def" layer being on an outer layer. > > claim 3. both positions must be used. > proof. suppose, to the contrary, that only one of these positions > is used, say with "abc" on an outer layer of the cube. > if the shape covers cell "f", then all three pieces cover > the central cell of the cube, a contradiction. on the other > hand, if the shape does not cover cell "f", then none of the > three pieces cover the central cell, again a contradiction. > > so now we know that one of the positions occurs twice and the other > occurs once. there is no loss of generality in supposing that "abc" > on the outer layer occurs twice, and "def" on the outer layer occurs > once. suppose that the shape has A cells of type "a" , B cells of > type "b", and so forth. now we count the number of corner cells, edge > cells, face cells and central cells of the 3x3x3 cube. this gives the > equations > > 2 A + D = 8 (counting corners) > A + 2 B + 2 D + E = 12 (counting edges) > B + 2 C + 2 E + F = 6 (counting faces) > C + 2 F = 1 (central cell) > > the last equation gives C = 1 and F = 0 , and then ... (hey, don't > we all know how to do these?) ... we find that the only solutions in > non-negative integers are > > A B C D E F > case 1: 4 4 1 0 0 0 > case 2: 3 2 1 2 1 0 > case 3: 2 0 1 4 2 0 > > we have three cases to analyze. > > in case 1, the shape is the 1x3x3 block. > > in case 2, consider the way the eight corners are covered. there are > three possibilities (up to symmetry) where the pieces are labeled > x, y and z. > > x.x ... y.y > situation 1: ... ... ... > x.z ... y.z > > > x.y ... y.y > situation 2: ... ... ... > x.z ... x.z > > > x.y ... z.y > situation 3: ... ... ... > x.y ... x.z > > in situation 1, piece z cannot reach the cell labeled t, so this cell > > x.x t.. y.y > ... ... ... > x.z ... y.z > > is filled by either piece x or piece y. without loss of generality, > suppose it is filled by piece x. the corresponding cell of piece y > must then also fill cell t, a contradiction. > > in situation 2, the bounding box of piece z is either parallel to the > bounding box of piece x, or parallel to the bounding box of piece y. > these two cases are equivalent by switching the roles of x and y. > so suppose the bounding box of piece z is as shown > > .zz .zz .zz > .zz .zz .zz > .zz .zz .zz > > then of the cells labeled t, 3 are covered by piece x (2 cells of > type "d" and 1 of type "e"), 6 are covered by piece z (3 of type "a", ) .t. .t. .t. ) .t. .t. .t. ) .t. .t. .t. > 2 of type "b" and 1 of type "c") and at least 1 is covered by piece y > (cell of type "c"). however, there are only 9 cells labeled t, > a contradiction. > > in situation 3, the cell labeled t cannot be filled by piece z, since > > x.y .v. z.y > ... v.v .t. > x.y .u. x.z > > it has no cells of type "f". now there is no loss of generality in > supposing it is filled by piece x. the corresponding cell of piece y > must be in cell t or cell u; cell t is occupied, so it must be in > cell u. of the cells labeled v, 1 is filled by piece x (cell of > type "c"), 1 is filled by piece y (cell of type "c"), and 2 are filled > by piece z (cells of type "b"). however, there are only 3 cells > labeled v, a contradiction. > > this finishes case 2. > > in case 3, consider the way the eight corners are covered. there are > two possibilities, up to symmetry > > z.z ... x.y > situation 1: ... ... ... > z.z ... y.x > > > z.z ... x.y > situation 2: ... ... ... > z.z ... x.y > > in situation 1, the cell labeled t must be filled by piece x, from > > z.z ... x.y > ... ... .t. > z.z ... y.x > > the cell of type "c". similarly, it must also be filled by piece y, > a contradiction. > > in situation 2, there are 3 possibilities (up to symmetry) for the > bounding boxes of pieces x and y: > > xx. xx. xx. .yy .yy .yy > subcase a: xx. xx. xx. .yy .yy .yy > xx. xx. xx. .yy .yy .yy > > > ... xxx xxx ... yyy yyy > subcase b: ... xxx xxx ... yyy yyy > ... xxx xxx ... yyy yyy > > > xx. xx. xx. ... yyy yyy > subcase c: xx. xx. xx. ... yyy yyy > xx. xx. xx. ... yyy yyy > > in subcase a, of the cells labeled t, 6 are covered by piece x, > > .t. .t. .t. > .t. .t. .t. > .t. .t. .t. > > and 6 are also covered by piece y, a contradiction. > > in subcase b, of the cells labeled t, 6 are covered by piece x, > > ... ttt ... > ... ttt ... > ... ttt ... > > and 6 are also covered by piece y, again a contradiction. > > in subcase c, of the cells labeled t, at least 1 is covered by > > ... ttt ... > ... ttt ... > ... ttt ... > > piece x, from its cell of type "c". 6 of these cells are covered > by piece y, and 3 are covered by piece z. there are only 9 cells > labeled t, so this is a contradiction. > > this finishes case 3, so we're done. > > mike > Return-Path: > Date: Sun, 29 Mar 1998 00:05:15 -0500 > From: michael reid > Reply-To: nobnet@iijnet.or.jp > Subject: [NOBNET 792] angus lavery's challenge refuted > Sender: owner-nobnet@iijnet.or.jp > To: nobnet@iijnet.or.jp > Errors-To: owner-nobnet@iijnet.or.jp > Message-Id: <199803290505.OAA08457@ml0.iijnet.or.jp> > X-Sequence: NOBNET 792 > Precedence: bulk > Lines: 28 > > rereading my post, i see that i omitted a diagram: > > (from case 2) > > > in situation 2, the bounding box of piece z is either parallel to the > > bounding box of piece x, or parallel to the bounding box of piece y. > > these two cases are equivalent by switching the roles of x and y. > > so suppose the bounding box of piece z is as shown > > > > .zz .zz .zz > > .zz .zz .zz > > .zz .zz .zz > > > > then of the cells labeled t, 3 are covered by piece x (2 cells of > > type "d" and 1 of type "e"), 6 are covered by piece z (3 of type "a", > > 2 of type "b" and 1 of type "c") and at least 1 is covered by piece y > > (cell of type "c"). however, there are only 9 cells labeled t, > > a contradiction. > > the missing diagram should be > > .t. .t. .t. > .t. .t. .t. > .t. .t. .t. > > which shows where the cells labeled t are. > > mike